smc

We know that $p^3+q^3+r^3=(p+q+r)(p^2+q^2+r^2-pq-qr-pr)+3pqr$. By Vieta's formulas, $p+q+r=1$,$pqr=2$, and $pq+qr+pr=1$. So if we can find $p^2+q^2+r^2$, we are done. Notice that $(p+q+r{)}^2=p^2+q^2+r^2+2pq+2qr+2pr$, so $p^2+q^2+r^2=(p+q+r{)}^2-2(pq+qr+pr)=1^2-2\cdot 1=-1$, which means that $p^3+q^3+r^3=1\cdot -2+3\cdot 2=\boxed{\text{none of these}}$